The bent wire shown in the figure lies in a uniform magnetic

The bent wire shown in the figure lies in a uniform magnetic field. Each straight section is 2.92 m long and makes an angle of theta = 75.0 degree with the x axis, and the wire carries a current of 2.56 A. What is the net magnetic force on the wire in unit-vector notation if the magnetic field is given by (a)4.70 k T? (b)4.17 i T?

Solution

For left section :

Length L = 2.92 cos 75 i + 2.92 sin 75 j

              = 0.7557 i +2.82 j

Current i = 2.56 A

Magnetic field B = 4.7 k

magnetic forlce on this segment F = i ( L X B )

                                                  = 2.56 [ (0.7557 i +2.82 j ) X 4.7 k ]

                                                  = 2.56 [ (0.7557 x4.7 )(i X k ) +(2.82 x4.7)(jXk)]

                                                  = 2.56 [ 3.551 (-j) +13.25 i ]

                                                  = - 9.092 j +33.92 i

Similarly ,

For right section :

Length L = 2.92 cos 75 i + 2.92 sin 75 (-j)

              = 0.7557 i - 2.82 j

Current i = 2.56 A

Magnetic field B = 4.7 k

magnetic forlce on this segment F \' = i ( L X B )

                                                  = 2.56 [ (0.7557 i -2.82 j ) X 4.7 k ]

                                                  = 2.56 [ (0.7557 x4.7 )(i X k ) +(-2.82 x4.7)(jXk)]

                                                  = 2.56 [ 3.551 (-j) -13.25 i ]

                                                  = - 9.092 j -33.92 i
Therefore net force F \" = F + F \'

                                 = (- 9.092 j +33.92 i)+(- 9.092 j -33.92 i)

                                 = -9.092 j -9.092 j

                                 = -18.184 j